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Inverse of adj A, or adj of A^-1

This type of question, asked 4 times: 21 Apr 2025 S2, 18 Apr 2026 S1, 19 Apr 2026 S2, 12 May 2026 S1. All 4 questions below, oldest first.

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  1. Question 1

    21 Apr 2025 S2

    108
    Let A be a non-singular matrix of order n and ∣A∣=k|A| =k, then (adjA)−1(adjA)^{- 1} is

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  2. Question 2

    18 Apr 2026 S1

    65
    Let A=[cos⁡α−sin⁡α0sin⁡αcos⁡α0001]A = \begin{bmatrix} \cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}. If B=adj AB = \mathrm{adj}\,A, then the matrix B−1B^{-1} is equal to...

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  3. Question 3

    19 Apr 2026 S2

    65
    If matrix A=[−12025202602202700−1]A = \begin{bmatrix} -1 & 2025 & 2026 \\ 0 & 2 & 2027 \\ 0 & 0 & -1 \end{bmatrix}, then the sum of all elements in adj(A−1)\mathrm{adj}(A^{-1}) is equal to...

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  4. Question 4

    12 May 2026 S1

    68
    Let AA be an invertible matrix. Identify the correct statement(s) from the following. (I) A(adj A−1)=1∣A∣A2A(\mathrm{adj}\,A^{-1}) = \dfrac{1}{|A|}A^2 (II) (adj A−1)−1=1∣A∣A−1(\mathrm{adj}\,A^{-1})^{-1} = \dfrac{1}{|A|}A^{-1} (III) (adj A)=1∣A∣A−1(\mathrm{adj}\,A) = \dfrac{1}{|A|}A^{-1}

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